If P ( n ) : 2 n < n ! , n ∈ ℕ , then P ( n ) is true for all n ≥ ____.

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t P ( n ) : 2 n < n ! ∀ n ∈ N F o r n = 1 2 < 1 ( N o t T r u e ) F o r n = 2 2 × 2 < 2 ! ⇒ 4 < 2 ( N o t T r u e ) F o r n = 3 2 × 3 < 3 ! ⇒ 6 < 3 . 2 . 1 ⇒ 6 < 6 ( N o t T r u e ) F o r n = 4 2 × 4 < 4 ! ⇒ 8 < 4 . 3 . 2 . 1 ⇒ 8 < 2 4 ( T r u e ) F o r n = 5 2 × 5 < 5 ! ⇒ 1 0 < 5 . 4 . 3 . 2 . 1 ⇒ 1 0 < 1 2 0 ( T r u e ) S o , P ( n ) i s t r u e f o r n ≥ 4 . H e n c e , t h e v a l u e o f t h e f i l l e r i s 4 .

Give an example of a statement P(n) which is true for all n≥4 but P(1) , P(2) , and P(3) are not true. Justify your answer.

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: T h e r e q u i r e d s t a t e m e n t i s P ( n ) = 2 n < n ! J u s t i f i c a t i o n : P ( n ) : 2 n < n ! P ( 1 ) : 2 . 1 < 1 ! ⇒ 2 < 1 n o t t r u e P ( 2 ) : 2 . 2 < 2 ! ⇒ 4 < 2 . 1 ⇒ 4 < 2 n o t t r u e P ( 3 ) : 2 . 3 < 3 ! ⇒ 6 < 3 . 2 . 1 ⇒ 6 < 6 n o t t r u e P ( 4 ) : 2 . 4 < 4 ! ⇒ 8 < 4 . 3 . 2 . 1 ⇒ 8 < 2 4 T r u e P ( 5 ) : 2 . 5 < 5 ! ⇒ 1 0 < 5 . 4 . 3 . 2 . 1 ⇒ 1 0 < 1 2 0 T r u e H e n c e , P ( n ) : 2 n < n ! i s n o t t r u e f o r P ( 1 ) , P ( 2 ) a n d P ( 3 ) b u t i t i s t r u e f o r a l l v a l u e s o f n ≥ 4 .

A sequence a1,a2,a3,… is defined by letting a1=3 and ak=7ak−1 , for all k≥2 . Show that an=3⋅7n−1 , for all n ∈ ℕ .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : a 1 = 3 a 2 = 7 a 2 − 1 = 7 . a 1 = 7 . 3 = 2 1 a 3 = 7 . a 3 − 1 = 7 . a 2 = 7 . 2 1 = 1 4 7 … L e t P ( n ) : a n = 3 . 7 n − 1 , ∀ n ∈ N . S t e p 1 : P ( 2 ) : a 2 = 3 . 7 2 − 1 = 2 1 ⇒ 2 1 = 2 1 w h i c h i s t r u e f o r P ( 2 ) . S t e p 2 : P ( k ) : a k = 3 . 7 k − 1 . L e t i t b e t r u e . S t e p 3 : a k = 7 a k − 1 ( G i v e n ) P u t k = k + 1 a k + 1 = 7 a k = 7 ( 3 . 7 k − 1 ) = 3 . 7 k + 1 − 1 = 3 . 7 ( k + 1 ) − 1 w h i c h i s t r u e f o r P ( k + 1 ) . H e n c e , P ( k + 1 ) i s t r u e w h e n e v e r P ( k ) i s t r u e .

A sequence d1,d2,d3,… is defined by letting d1=2 and dk=dk−1k , for all k≥2 . Show that dn=2n! , for all n ∈ ℕ .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : d 1 = 2 a n d d k = d k − 1 k L e t P ( n ) : d n = 2 n ! S t e p 1 : P ( 1 ) : d 1 = 2 1 ! = 2 w h i c h i s t r u e f o r P ( 1 ) . S t e p 2 : P ( k ) : d k = 2 k ! . L e t i t b e t r u e f o r P ( k ) . S t e p 3 : G i v e n t h a t : d k = d k − 1 k ∴ d k + 1 = d k + 1 − 1 k + 1 = d k k + 1 ⇒ d k + 1 = 1 k + 1 . d k = 1 k + 1 . 2 k ! ⇒ d k + 1 = 2 ( k + 1 ) ! w h i c h i s t r u e f o r P ( k + 1 ) . H e n c e , P ( k + 1 ) i s t r u e w h e n e v e r P ( k ) i s t r u e .

A sequence b0,b1,b2,… is defined by letting b0=5 and bk=4+bk−1 , for all k≥1 . Show that bn=5+4n , for all n ∈ ℕ .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: W e h a v e b 0 = 5 a n d b k = 4 + b k − 1 ⇒ b 0 = 5 , b 1 = 4 + b 0 = 4 + 5 = 9 a n d b 2 = 4 + b 1 = 4 + 9 = 1 3 L e t P ( n ) : b n = 5 + 4 n S t e p 1 : P ( 1 ) : b 1 = 5 + 4 = 9 ⇒ 9 = 9 w h i c h i s t r u e f o r P ( 1 ) . S t e p 2 : P ( k ) : b k = 5 + 4 k . L e t i t b e t r u e ∀ k ∈ N . S t e p 3 : G i v e n t h a t : P ( k ) = 4 + b k − 1 ⇒ P ( k + 1 ) = 4 + b k + 1 − 1 ⇒ P ( k + 1 ) = 4 + b k = 4 + 5 + 4 k ⇒ P ( k + 1 ) = 4 + b k = 5 + 4 ( k + 1 ) w h i c h i s t r u e f o r P ( k + 1 ) . H e n c e , P ( k + 1 ) i s t r u e w h e n e v e r P ( k ) i s t r u e .

Prove that, for all n ∈ ℕ : c o s α + c o s ( α + β ) + c o s ( α + 2 β ) + ⋯ + c o s ( α + ( n − 1 ) β ) = c o s ( α + ( n − 1 ) β 2 ) s i n n β 2 s i n β 2 .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: L e t P ( n ) : c o s α + c o s ( α + β ) + c o s ( α + 2 β ) + … + c o s [ α + ( n − 1 ) β ] = c o s [ α + ( n − 1 2 ) β ] [ s i n n β 2 ] s i n β 2 S t e p 1 : P ( 1 ) : c o s α = ( c o s α ) ( s i n β 2 ) s i n β 2 = c o s α w h i c h i s t r u e f o r P ( 1 ) . S t e p 2 : P ( k ) : c o s α + c o s ( α + β ) + c o s ( α + 2 β ) + … + c o s [ α + ( k − 1 ) β ] = c o s [ α + ( k − 1 2 ) β ] [ s i n k β 2 ] s i n β 2 L e t i t b e t r u e . S t e p 3 : P ( k + 1 ) : c o s α + c o s ( α + β ) + c o s ( α + 2 β ) + … + c o s [ α + ( k − 1 ) β ] + c o s [ α + ( k + 1 − 1 ) β ] = c o s [ α + ( k − 1 2 ) β ] [ s i n k β 2 ] s i n β 2 + c o s ( α + k β ) [ F r o m S t e p 2 ] = 2 c o s [ α + ( k − 1 2 ) β ] ( s i n k β 2 ) + 2 c o s ( α + k β ) . s i n β 2 2 s i n β 2 = s i n [ α + k β − β 2 ] − s i n [ α − β 2 ] + s i n [ α + k β + β 2 ] − s i n [ α + k β − β 2 ] 2 s i n β 2 [ ? s i n A − s i n B = 2 c o s A + B 2 . s i n A − B 2 ] = c o s ( α + k β 2 ) . s i n ( k + 1 ) β 2 s i n β 2 = c o s [ α + ( k + 1 − 1 2 ) β ] . s i n ( k + 1 2 ) β s i n β 2 w h i c h i s t r u e f o r P ( k + 1 ) . H e n c e , P ( k + 1 ) i s t r u e w h e n e v e r P ( k ) i s t r u e .

Prove that c o s θ ⋅ c o s 2 θ ⋅ c o s 4 θ ⋯ c o s 2 n − 1 θ = s i n ( 2 n θ ) 2 n s i n θ , for all n ∈ ℕ .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: L e t P ( n ) : c o s θ . c o s 2 θ . c o s 2 2 θ + … + c o s 2 n − 1 θ = s i n 2 n θ 2 n s i n θ , f o r a l l n ∈ N . S t e p 1 : P ( 1 ) : c o s θ = s i n 2 1 θ 2 1 s i n θ = s i n 2 θ 2 s i n θ = 2 s i n θ . c o s θ 2 s i n θ = c o s θ ⇒ c o s θ = c o s θ w h i c h i s t r u e f o r P ( 1 ) . S t e p 2 : P ( k ) : c o s θ . c o s 2 θ . c o s 2 2 θ + … + c o s 2 k − 1 θ = s i n 2 k θ 2 k s i n θ L e t i t b e t r u e f o r P ( k ) . S t e p 3 : P ( k + 1 ) : c o s θ . c o s 2 θ . c o s 2 2 θ + … + c o s 2 k − 1 θ . c o s 2 ( k + 1 ) − 1 θ = s i n 2 k θ 2 k s i n θ . c o s 2 ( k + 1 ) − 1 θ = s i n 2 k θ 2 k s i n θ . c o s 2 k θ = 2 s i n 2 k θ . c o s 2 k θ 2 . 2 k s i n θ = s i n 2 . 2 k θ 2 k + 1 s i n θ [ ? 2 s i n θ c o s θ = s i n 2 θ ] = s i n 2 k + 1 θ 2 k + 1 s i n θ w h i c h i s t r u e f o r P ( k + 1 ) . H e n c e , P ( k + 1 ) i s t r u e w h e n e v e r P ( k ) i s t r u e .

Let E? denote the complement of an event E. Let E?, E?, and E? be any pairwise independent events with P(E?)>0 and P(E?∩E?∩E?)=0. Then P(E??∩E??/E?) is equal to

P(E₃ᶜ) - P(E₂ᶜ)

P(E₂ᶜ) + P(E₃)

The set of all possible values of θ in the interval (0,π) for which the points (1,2) and (sinθ,cosθ) lie on the same side of the line x+y=1 is

(0, π/4)

(0, π/2)

(0, 3π/4)

The area (in sq. units) of an equilateral triangle inscribed in the parabola y²=8x, with one of its vertices on the vertex of this parabola, is

256√3

Maths NCERT Exemplar Solutions Class 11th Chapter Four: Overview, Questions, Preparation

Q: If 1 0 n + 3 ⋅ 4 n + 2 + k is divisible by 9 for all n ∈ ℕ , then the least positive integral value of k is: (a) 5 (b) 3 (c) 7 (d) 1

This is a Objective Type Questions as classified in NCERT Exemplar Sol: L e t P ( n ) = 1 0 n + 3 . 4 n + 2 + k i s d i v i s i b l e b y 9 , ∀ n ∈ N P ( 1 ) = 1 0 1 + 3 . 4 1 + 2 + k = 1 0 + 3 . 6 4 + k = 1 0 + 1 9 2 + k = 2 0 2 + k m u s t b e d i v i s i b l e b y 9 . I f ( 2 0 2 + k ) i s d i v i s i b l e b y 9 t h e n k m u s t b e e q u a l t o 5 2 0 2 + 5 = 2 0 7 w h i c h i s d i v i s i b l e b y 9 = 2 0 7 9 = 2 3 So, the least positive integral value of k=5. H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

Q: For all n ∈ ℕ , 3 ⋅ 5 2 n + 1 + 2 ⋅ 3 n + 1 is divisible by: (a) 19 (b) 17 (c) 23 (d) 25

This is a Objective Type Questions as classified in NCERT Exemplar Sol: L e t P ( n ) = 3 . 5 2 n + 1 + 2 3 n + 1 P ( 1 ) = 3 . 5 2 . 1 + 1 + 2 3 . 1 + 1 = 3 . 5 3 + 2 4 = 3 ( 1 2 5 ) + 1 6 = 3 7 5 + 1 6 = 3 9 1 = 2 3 × 1 7 S o , i t i s d i v i s i b l e b y 1 7 a n d 2 3 b o t h . H e n c e , t h e c o r r e c t o p t i o n i s ( b ) a n d ( c ) .

Q: If x n − 1 is divisible by x − k , then the least positive integral value of k is: (a) 1 (b) 2 (c) 3 (d) 4

This is a Objective Type Questions as classified in NCERT Exemplar Sol: L e t P ( n ) = x n − 1 i s d i v i s i b l e b y x − k . P ( 1 ) = x − 1 i s d i v i s i b l e b y x − k . Since k=1 is the possible least integral value of k. H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

Q: If P ( n ) : 2 n < n ! , n ∈ ℕ , then P ( n ) is true for all n ≥ ____.

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t P ( n ) : 2 n < n ! ∀ n ∈ N F o r n = 1 2 < 1 ( N o t T r u e ) F o r n = 2 2 × 2 < 2 ! ⇒ 4 < 2 ( N o t T r u e ) F o r n = 3 2 × 3 < 3 ! ⇒ 6 < 3 . 2 . 1 ⇒ 6 < 6 ( N o t T r u e ) F o r n = 4 2 × 4 < 4 ! ⇒ 8 < 4 . 3 . 2 . 1 ⇒ 8 < 2 4 ( T r u e ) F o r n = 5 2 × 5 < 5 ! ⇒ 1 0 < 5 . 4 . 3 . 2 . 1 ⇒ 1 0 < 1 2 0 ( T r u e ) S o , P ( n ) i s t r u e f o r n ≥ 4 . H e n c e , t h e v a l u e o f t h e f i l l e r i s 4 .

Q: Let P(n) be a statement and let P(k + 1) for some natural number k . Then P(n) is true for all n ∈ ℕ .

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t : P ( k ) ⇒ P ( k + 1 ) P ( 1 ) ⇒ P ( 2 ) w h i c h i s n o t t r u e . H e n c e , t h e s t a t e m e n t i s ' F a l s e ' .

Updated on Jul 25, 2025 11:40 IST

By Raj Pandey

Table of content

Principle of Mathematical Induction Long Answer Type Questions
Principle of Mathematical Induction Short Answer Type Questions
Principle of Mathematical Induction Objective Type Questions
Jee mains

Principle of Mathematical Induction Short Answer Type Questions

Give an example of a statement $P (n)$ which is true for all $n \geq 4$ but $P (1)$ , $P (2)$ , and $P (3)$ are not true. Justify your answer.

\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) = 2 n < n! \\ J u s t i f i c a t i o n : P (n) : 2 n < n! \\ P (1) : 2.1 < 1! \Rightarrow 2 < 1 n o t t r u e \\ P (2) : 2.2 < 2! \Rightarrow 4 < 2.1 \Rightarrow 4 < 2 n o t t r u e \\ P (3) : 2.3 < 3! \Rightarrow 6 < 3.2.1 \Rightarrow 6 < 6 n o t t r u e \\ P (4) : 2.4 < 4! \Rightarrow 8 < 4.3.2.1 \Rightarrow 8 < 24 T r u e \\ P (5) : 2.5 < 5! \Rightarrow 10 < 5.4.3.2.1 \Rightarrow 10 < 120 T r u e \\ H e n c e, P (n) : 2 n < n! i s n o t t r u e f o r P (1), P (2) a n d P (3) b u t i t i s t r u e f o r a l l v a l u e s o f n \geq 4 . \end{array}

Give an example of a statement $P (n)$ which is true for all $n$ . Justify your answer.

\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) : 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ J u s t i f i c a t i o n : P (1) : 1 = \frac{1 (1 + 1)}{2} \\ P (k) : 1 + 2 + 3 + \dots + k = \frac{k (k + 1)}{2} . L e t i t b e t r u e . \\ P (k + 1) : 1 + 2 + 3 + \dots + k + (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{(k + 1) (k + 2)}{2} \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}

Commonly asked questions

Give an example of a statement $P (n)$ which is true for all $n \geq 4$ but $P (1)$ , $P (2)$ , and $P (3)$ are not true. Justify your answer.